Problem: You have found the following ages (in years) of all 5 snakes at your local zoo: $ 22,\enspace 10,\enspace 22,\enspace 20,\enspace 8$ What is the average age of the snakes at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 5 snakes at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{22 + 10 + 22 + 20 + 8}{{5}} = {16.4\text{ years old}} $ Find the squared deviations from the mean for each snake. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $22$ years $5.6$ years $31.36$ years $^2$ $10$ years $-6.4$ years $40.96$ years $^2$ $22$ years $5.6$ years $31.36$ years $^2$ $20$ years $3.6$ years $12.96$ years $^2$ $8$ years $-8.4$ years $70.56$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{31.36} + {40.96} + {31.36} + {12.96} + {70.56}} {{5}} $ $ {\sigma^2} = \dfrac{{187.2}}{{5}} = {37.44\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{37.44\text{ years}^2}} = {6.1\text{ years}} $ The average snake at the zoo is 16.4 years old. There is a standard deviation of 6.1 years.